displaystyle limx → 0 (cos x+sin x)(1/x) equals

Question

displaystyle limx → 0 (cos x+sin x)(1/x) equals (A) e1 (B) e-1 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

displaystyle limx → 0 (cos x+sin x)(1/x) (1∞ form) =e displaystyle limx → 0((cos x+sin x-1)/x) =e displaystyle limx → 0[(sin x/x)-((1-cos x/x))] =e1-0=e1

Q. $x \to 0 lim$ $(cos x + s in x)^{\frac{1}{x}}$ equals

$e^{1}$

$e^{- 1}$

$1$

$0$

$x \to 0 lim$ $(cos x + s in x)^{\frac{1}{x}} (1^{\infty}$ form $)$
$= e^{x \to 0 lim \frac{( cos x + s in x - 1 )}{x}}$
$= e^{x \to 0 lim [\frac{s in x}{x} - (\frac{1 - cos x}{x})]}$
$= e^{1 - 0} = e^{1}$

Q. x→0lim​ (cosx+sinx)x1​ equals

e1

e−1

1

0

x→0lim​ (cosx+sinx)x1​(1∞ form) =ex→0lim​x(cosx+sinx−1)​ =ex→0lim​[xsinx​−(x1−cosx​)] =e1−0=e1

Q. $x \to 0 lim$ $(cos x + s in x)^{\frac{1}{x}}$ equals

$e^{1}$

$e^{- 1}$

$1$

$0$

$x \to 0 lim$ $(cos x + s in x)^{\frac{1}{x}} (1^{\infty}$ form $)$
$= e^{x \to 0 lim \frac{( cos x + s in x - 1 )}{x}}$
$= e^{x \to 0 lim [\frac{s in x}{x} - (\frac{1 - cos x}{x})]}$
$= e^{1 - 0} = e^{1}$